Question

Urgentt ! Dau coroana ! Cât H si O sunt necesari pt a obține 1 kg de apa. Calculati in grame si in moli

Answer 1

$m_{H_2O}=1 \ kg=1 \ 000 \ g \\ \boxed{m_{H_2}-?} \\ \boxed{\nu_{H_2}-?} \\ \boxed{m_{O_2}-?} \\ \boxed{\nu_{O_2}-?} \\ \bold{Rezolvare:} \\ 2H_2+O_2=2H_2O \\ \boxed{\nu=\frac{m}{M}} \\ \nu_{H_2}=\nu_{H_2O} \\ \nu_{H_2O}=\frac{m_{H_2O}}{M_{H_2O}} \\ M_{H_2O}=1 \cdot 2+16 \cdot 1=2+16=18 \ g/mol \\ \nu_{H_2O}=\frac{1 \ 000 \ g}{18 \ g/mol}=55,6 \ moli \\ \Rightarrow \boxed{\bold{\nu_{H_2}=55,6 \ moli}} \\ m_{H_2}=\nu_{H_2} \cdot M_{H_2} \\ m_{H_2}=1 \cdot 2=2 \ g/mol \\ m_{H_2}=55,6 \ moli \cdot 2 \ g/mol$
$\Rightarrow \boxed{\bold{m_{H_2}=111,2 \ g}} \\ \frac{\nu_{O_2}}{1 \ mol}=\frac{\nu_{H_2O}}{2\ moli}=\frac{55,6 \ moli}{2 \ moli} \\ \nu_{O_2}=\frac{55,6 \ moli \cdot 1 \ mol}{2 \ moli}=\frac{55,6 \ moli}{2 \ moli} \\ \Rightarrow \boxed{\bold{\nu_{O_2}=27,8 \ moli}} \\ m_{O_2}=\nu_{O_2} \cdot M_{O_2} \\ M_{O_2}=16 \cdot 2=32 \ g/mol \\ \nu_{O_2}=27,8 \ moli \cdot 32 \ g/mol \\ \Rightarrow \boxed{\bold{m_{O_2}=889,6 \ g}}$